# Time

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Time is divided different time units in Kerbal Space Program. Since 0.23.5 those time units base on Kerbin instead of Earth. So each day has the length of the Kerbin day[specify: solar or sidereal?] and each year has the length of one Kerbin year. Because both are considerably shorter than Earth's equivalents both will be longer. The smaller time units hours and minutes have the same length like the Earth's hour and minute.

## Kerbin calendar

The following table outlines the orbital period and frequency of each celestial body in the Kerbol System. The following table calculates Earth Months as 365 days/12 months/year ≈ 30.4 days. A sidereal Kerbin day is 6 hours long, the Mun has an orbital period of 38.6 hours which defines a Kerbin month, and Kerbin has an orbital period of 2556.5 hours which defines a Kerbin year.

Celestial Body Parent Hours Kerbin Earth Revolutions
per Hour
Revolutions per Kerbin Revolutions per Earth
Days Months Years Days Months Years Day Month Year Day Month Year
Moho Kerbol 615.49 102.58 15.95 0.24 25.65 0.84 0.07 0.00 0.01 0.06 4.15 0.04 1.19 14.23
Eve Kerbol 1571.7 261.95 40.72 0.61 65.49 2.15 0.18 0.00 0.00 0.02 1.63 0.02 0.46 5.57
Gilly Eve 107.9 17.98 2.80 0.04 4.50 0.15 0.01 0.01 0.06 0.36 23.69 0.22 6.77 81.19
Kerbin Kerbol 2556.50 426.08 66.23 1.00 106.52 3.50 0.29 0.00 0.00 0.02 1.00 0.01 0.29 3.43
Mun Kerbin 38.60 6.43 1.00 0.02 1.61 0.05 0.00 0.03 0.16 1.00 66.23 0.62 18.91 226.94
Minmus Kerbin 299.50 49.92 7.76 0.12 12.48 0.41 0.03 0.00 0.02 0.13 8.54 0.08 2.44 29.25
Duna Kerbol 4809.80 801.63 124.61 1.88 200.41 6.59 0.55 0.00 0.00 0.01 0.53 0.00 0.15 1.82
Ike Duna 18.20 3.03 0.47 0.01 0.76 0.02 0.00 0.05 0.33 2.12 140.47 1.32 40.11 481.32
Dres Kerbol 13303.60 2217.27 344.65 5.20 554.32 18.22 1.52 0.00 0.00 0.00 0.19 0.00 0.05 0.66
Jool Kerbol 29072.60 4845.43 753.18 11.37 1211.36 39.83 3.32 0.00 0.00 0.00 0.09 0.00 0.03 0.30
Laythe Jool 14.70 2.45 0.38 0.01 0.61 0.02 0.00 0.07 0.41 2.63 173.91 1.63 49.66 595.92
Vall Jool 29.43 4.91 0.76 0.01 1.23 0.04 0.00 0.03 0.20 1.31 86.87 0.82 24.80 297.66
Tylo Jool 58.87 9.81 1.53 0.02 2.45 0.08 0.01 0.02 0.10 0.66 43.43 0.41 12.40 148.80
Bop Jool 110.92 18.49 2.87 0.04 4.62 0.15 0.01 0.01 0.05 0.35 23.05 0.22 6.58 78.98
Pol Jool 153.70 25.62 3.98 0.06 6.40 0.21 0.02 0.01 0.04 0.25 16.63 0.16 4.75 56.99
Eeloo Kerbol 43608.90 7268.15 1129.76 17.06 1817.04 59.74 4.98 0.00 0.00 0.00 0.06 0.00 0.02 0.20

## Phase angles

The phase angle determine the angular distance between two bodies around the same object. The phase angle of a body exactly on the other side of the central body is 180°. If the central body is on one side, and the two compared bodies are in line, the phase angle is 0°.

With the starting phase angles listed below it is possible to calculate the phase angle at any given moment. This information can be used to determine if both bodies are aligned for a transfer orbit. To calculate this value, the revolutions per second (RPS) for both bodies is required. This is the inverse of the orbital period:

${\displaystyle r={\frac {1}{p}}}$
Where:
• ${\displaystyle r}$ is the revolutions per time unit
• ${\displaystyle p}$ is the orbital period in the time unit
Both time units are second for RPS

The phase angle change per second is then simply the subtraction of the origin from the target's planet RPS. This also tells when the current alignment repeat again, by inverting the phase angle change again. This time is called synodic orbital period.

${\displaystyle p_{\text{synodic}}=\left\vert {\frac {1}{{\frac {1}{p_{t}}}-{\frac {1}{p_{o}}}}}\right\vert }$
Where:
• ${\displaystyle p_{\text{synodic}}}$ is the synodic orbital period
• ${\displaystyle p_{t}}$ is the sidereal orbital period of the targeted body
• ${\displaystyle p_{o}}$ is the sidereal orbital period of the original body

There are calculation tools available online which tell which is the best phase angle for an efficient transfer. Knowing this, the first alignment date can be calculated by first subtracting the starting phase angle from the desired phase angle. This gives the required first change with needs to be divided by the the phase angle change to determine how long it takes to get the first alignment.

### Starting phase angles

The following table shows the phase angle in degrees and revolutions of each planet at the beginning of each game (UT = 0s).

Planet Degrees Revolutions
Moho 84.92° 0.23589
Eve 15.00° 0.04167
Duna 135.51° 0.37642
Jool 238.43° 0.66231
Dres 10.02° 0.02783
Eeloo 309.98° 0.86106

### Example

The revolutions per second for Duna is:

${\displaystyle r={\frac {1}{17315400s}}=5.77520589\cdot 10^{-8}Hz}$

The phase angle change per second relative to Kerbin is then:

${\displaystyle r_{\text{relative}}=r_{\text{Duna}}-r_{\text{Kerbin}}=-5.09017\cdot 10^{-8}Hz}$

The optimal phase angle for a transfer from Kerbin to Duna is 0.12323 revolutions or 44.3628°. The starting phase angle is 0.37642 revolutions so the first time change required is -0.25319 revolutions.

${\displaystyle t_{0}={\frac {\beta }{r_{\text{relative}}}}={\frac {\beta }{r_{\text{Duna}}-r_{\text{Kerbin}}}}={\frac {-0.25319}{-5.09017\cdot 10^{-8}Hz}}=4974119s}$

The transfer windows will then repeat periodically using the synodic period between both bodies:

${\displaystyle p_{\text{synodic}}=\left\vert {\frac {1}{{\frac {1}{p_{\text{Duna}}}}-{\frac {1}{p_{\text{Kerbin}}}}}}\right\vert =\left\vert {\frac {1}{-5.09017\cdot 10^{-8}Hz}}\right\vert =19645709s}$
${\displaystyle t_{n}=t_{0}+n\cdot p_{\text{synodic}}}$